What is the integral of (x^3+x^2+2)/(x^2-x) ?

The integral of (x^3+x^2+2)/(x^2-x) is 1/2 x^2+2x+ln(2|x-1|)+3ln|x-1|-2ln|x|-3/2-1+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of (x^3+x^2+2)/(x^2-x)

\int \frac{x ^{3} + x ^{2} + 2}{x ^{2} - x} d x

\frac{1}{2} x^{2} + 2 x + ln (2 ∣ x - 1 ∣) + 3 ln ∣ x - 1 ∣ - 2 ln ∣ x ∣ - \frac{3}{2} - 1 + C

Solution steps

\int \frac{x ^{3} + x ^{2} + 2}{x ^{2} - x} d x

Expand

\frac{x ^{3} + x ^{2} + 2}{x ^{2} - x} : \frac{x ^{3}}{x ^{2} - x} + \frac{x ^{2}}{x ^{2} - x} + \frac{2}{x ^{2} - x}

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \int \frac{x ^{3}}{x ^{2} - x} d x + \int \frac{x ^{2}}{x ^{2} - x} d x + \int \frac{2}{x ^{2} - x} d x

\int \frac{x ^{3}}{x ^{2} - x} d x = \frac{1}{2} (2 ln (2 ∣ x - 1 ∣) + (x - 1) (x + 3))

\int \frac{x ^{2}}{x ^{2} - x} d x = x - 1 + ln ∣ x - 1 ∣

\int \frac{2}{x ^{2} - x} d x = 2 (- ln ∣ x ∣ + ln ∣ x - 1 ∣)

= \frac{1}{2} (2 ln (2 ∣ x - 1 ∣) + (x - 1) (x + 3)) + x - 1 + ln ∣ x - 1 ∣ + 2 (- ln ∣ x ∣ + ln ∣ x - 1 ∣)

Simplify

\frac{1}{2} (2 ln (2 ∣ x - 1 ∣) + (x - 1) (x + 3)) + x - 1 + ln ∣ x - 1 ∣ + 2 (- ln ∣ x ∣ + ln ∣ x - 1 ∣) : \frac{1}{2} x^{2} + 2 x + ln (2 ∣ x - 1 ∣) + 3 ln ∣ x - 1 ∣ - 2 ln ∣ x ∣ - \frac{3}{2} - 1

= \frac{1}{2} x^{2} + 2 x + ln (2 ∣ x - 1 ∣) + 3 ln ∣ x - 1 ∣ - 2 ln ∣ x ∣ - \frac{3}{2} - 1

Add a constant to the solution

= \frac{1}{2} x^{2} + 2 x + ln (2 ∣ x - 1 ∣) + 3 ln ∣ x - 1 ∣ - 2 ln ∣ x ∣ - \frac{3}{2} - 1 + C

\int \frac{1}{( x ^{2} + 2 x + 3 ) ^{\frac{3}{2}}} d x

\frac{d y}{d x} = 6 y x

x \to 0 lim (\frac{5}{x})

\frac{d y}{d x} + 5 e^{x + y} = 0

\int_{1}^{2} (2 x^{2}) - (2 x + 6) d x

What is the integral of (x^3+x^2+2)/(x^2-x) ?
The integral of (x^3+x^2+2)/(x^2-x) is 1/2 x^2+2x+ln(2|x-1|)+3ln|x-1|-2ln|x|-3/2-1+C